Question

Solve the following pair of linear equations for $x$ and $y$:

$px+qy=r$ and $qx-py=1+r$

• A. $x=\frac{q+r(p+q)}{p^2+q^2};y=\frac{r(q-p)-p}{p^2+q^2}$
• B. $x=\frac{r+p(q-r)}{q^2+p^2};y=\frac{q(r-p)-p}{q^2+p^2}$
• C. $x=\frac{qr+p(r+q)}{r^2+q^2};y=\frac{r(q-p)-p}{p^2+q^2}$
• D. $x=\frac{q-r(p+rq)}{q^2+{pq}^2};y=\frac{q(q-p)-r}{p^2+q^2}$

Answer 1

The correct option is A $x=\frac{q+r(p+q)}{p^2+q^2};y=\frac{r(q-p)-p}{p^2+q^2}$

Given equations are,

$px+qy=r$ ....(1)

and $qx-py=1+r$ ....(2)

Multiply equation (1) by $p$ and equation (2) by $q$,

$p^{2}x+pqy=pr$ ....(3)

and $q^{2}x-pqy=q+qr$ ....(4)

By adding equations (3) and (4), we get

$p^{2}x+q^{2}x=q+pr+qr$

$x=\frac{q+r(p+q)}{p^2+q^2}$

Put this value in equation (1), we get

$p\left(\frac{q+r(p+q)}{p^2+q^2}\right)+qy=r$

$\Rightarrow pq+p^{2}r+pqr+p^{2}qy+q^{3}y=p^{2}r+q^{2}r$

$\Rightarrow yq(p^2+q^2)=q^{2}r-pqr-pq$

$\Rightarrow y=\frac{qr-pr-p}{p^2+q^2}$

$\Rightarrow y=\frac{r(q-p)-p}{p^2+q^2}$